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Proof step 19 from proof BertrandsPostulate (previous step)

all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0, 4 ^ i ≤ 4 ^ ((2 * i) / 3) * (2 * i) ^ sqrt(2 * i), 0 < i, 2 < i, binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3), 4 ^ i ≤ binom_coef(2 * i, i) * 2 * i, i < 1024, ∀ j. j ∈ ints0 → (2 * i) / 3 < j ∧ j ≤ i, ∀ j. j ∈ ints1 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3, ∀ j. j ∈ ints2 → j ≤ sqrt(2 * i) ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

User applied rule insert lemma

And obtained the following new goals

Proof step 20:
⊦ ∀ i. 0 < i ∧ i ≤ 2 ^ 10 → (∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i)
Proof step 21:
∀ i. 0 < i ∧ i ≤ 2 ^ 10 → (∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i), all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0, 4 ^ i ≤ 4 ^ ((2 * i) / 3) * (2 * i) ^ sqrt(2 * i), 0 < i, 2 < i, binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3), 4 ^ i ≤ binom_coef(2 * i, i) * 2 * i, i < 1024, ∀ j. j ∈ ints0 → (2 * i) / 3 < j ∧ j ≤ i, ∀ j. j ∈ ints1 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3, ∀ j. j ∈ ints2 → j ≤ sqrt(2 * i) ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

Rule argument

spec: inst: name: bertrand-holds-lower seq: ⊦ ∀ i. 0 < i ∧ i ≤ 2 ^ 10 → (∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i) subst: flags: current, []

Additional information

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